 Hello everyone, I welcome you all to MSB lecture series on interpretive spectroscopy. So now I am discussing about UVS spectroscopy. In my last couple of lectures, I was discussing about Orgel diagrams, combined Orgel diagrams for two sets of D electronic configurations. The first one was about D1, D4, D6, D9 for both octahedral and tetrahedral complexes. With one Orgel diagram for high spin complexes, we should be able to interpret data obtained from trans spectroscopy. And in the same way, another Orgel diagram I showed you, that is for D2, D3 and D7, D8 system for both octahedral and tetrahedral complexes combined together. And also we saw some anomalies, energy of some transits are decreasing and the energy of some transits are increasing. How you can correct that one using RACA parameters, considering nephelacetic effects. And now we will consider another unique electronic configuration, D5. So spectroph D5 ions, high spin one, we have quite a few examples are there, manganese two complexes and iron three complexes are D5 system. And for example, if you say hexafluoromanganate 4 minus or hexaqua manganese 2 plus or hexafluoroferrate 3 minus, all are having D5 electronic configuration and high spin complexes. And if you consider electronic rearrangement in this high spin complexes and if you recall spin selection rule of delta S equals 0, these transits are forbidden because if any electron you promote, they will be going with upward spin, then you will be having two spins with the same spin value of plus half. In that case it is not allowed, that is the reason they are all spin forbidden transitions. The ground term is 6S with 11 excited states here. So that means, transient probabilities are extremely low and here in order to see transition from D5 system that is 6S to 4G, 4F, 4D, 4P that involves the reversal of one spin. That means, if I promote one of the electron, it should go that means, it is basically reversal of one spin is essential to see the transition from this one to this four. On the other hand, in order to see transition to these seven levels, we need to reverse the spin of both the electrons that is doubly spin forbidden. So, in this case what happens, it is very weak. If you look into the spectrum, so this how the spectrum looks like for hexaqua manganese 2 plus. So, manganese 2 has a D5 high spin electronic configuration, all derbs are occupied with one electron each. So, none of the possible DD transition is spin allowed. Since for any transition, the spin of the electron must be reversal, both higher energy EGR builds contain already one electron according to Pauli's principle, the spin of the second electron must be reversal. So, therefore, all possible transitions are very weak and hence hexaqua manganese 2 plus is very very pale in color. The bands are extremely weak that is reflected in its epsilon value of 0.2 to 0.03 liter per mole per centimeter. So, allowed transitions, spin allowed bands are invariably broad. So, you can see here these are the transitions observed for hexaqua manganese 2 plus. These arrows indicate the position of the predicted band positions here. So, now the Argel diagram, if you write for this unique D5 electronic configuration for both tetrahedral and octahedral, of course, in case of tetrahedral ignore G. Now, one can write all possible with ground state being 6A1G and 6S is 6A1G and now from 6A1G, you can see about 6 transitions the corresponding lambda maximum values are shown here. That is what I showed you in the spectrum here, this spectrum you can identify those transitions shown here. So, now let us come back to Raqqha parameters, Raqqha parameters were generated as a means to describe the effect of electron-electron repulsion within the metal complexes and the Raqqha parameters are A, B and C. In the case of Tanube-Sugunov diagrams, I will tell you what is Tanube-Sugunov diagrams. We learned about Argel diagrams, we have another set of diagrams which are called Tanube-Sugunov diagrams. In the case of Tanube-Sugunov diagrams, each electron configuration split has an energy that can be related by the B value. A is ignored because it is roughly the same for any metal center and C generally approximately as being 1 by 4 times B. What B represents is an approximation of the bond strength between the ligand and the metal. So, comparisons between tabulated free ion B and B of a coordinated complex is called the nephlaxitic ratio, the effect of reducing electron-electron repulsion via ligands that is beta equals beta complex over beta free ion. Now, let us come back to Tanube-Sugunov diagrams, so Tanube-Sugunov diagrams are used in coordination chemistry to predict electromagnetic absorption of metal coordination compounds of both tetrahedral and octahedral complexes. And ground state is always taken as apsis or x-axis or horizontal axis and provides a constant reference point, other energy levels are plotted relative to this ground state which is as same as x-axis. The low spin terms that is states where the spin multiple is to a spin is lower than the ground states are also included in this Tanube-Sugunov diagram whereas, those things are not considered in case of Argyle diagrams. In order to make the diagram general for different metal ions with the same electronic configuration and to allow for different ligands of different ligand field strength both of which affect DQ that is B and B prime the axes are plotted in units of energy by B and DQ by B. By doing this one in one Tanube-Sugunov diagram we can consider the entire band of ligands we come across in the spectrochemical series. What is the difference between Tanube-Sugunov diagram and Argyle diagram is different diagram is required for different electronic arrangement that means every electronic arrangement you need a separate Tanube-Sugunov diagram you need D1 a separate D2 separate D9 all D1 to D9 except D5 you need a separate Tanube-Sugunov diagram. But it includes all ligands having different ligand field strength the axis in a Tanube-Sugunov diagram is in terms of crystal field splitting parameter 10 DQ or delta octahedral scaled by the Birakha parameter this the y axis in terms of the energy of electronic transition E scaled by B. So, diagrams for D4, D5, D6, D7 metal ions have a discontinuity in energies as the ligand field is varied the discontinuity shown with the vertical line represents complexes changing from high spin to low spin complexes to the left of the line metal complexes are high spin as the spin pairing energy is greater than that of the ligand field splitting to the right of the line metal complexes are low spin as the spin pairing energy is less than that of the ligand field energy. So, what you can do is take Tanube-Sugunov diagrams for each electronic configuration and then read this paragraph here and then observe you can make out the differences how it looks like. So, you can understand in a better way. So, for example, here I have given for 6 D6 system here D2 with replace no fundamental difference between strong and weak field ligands and D6 cobalt 3 place if you can this discontinuity at the 10 DQ B equals 20, 10 DQ by B equals 20 at this point pairing of electron occurs. That means, basically we are moving from weak field to the strong field as a result what happens pairing starts to the left we have high spin complexes weak field to the right we have low spin complexes strong field ligand. So, that means, free ion ground state is 5 D in the octahedral field a singlet 1 I of high energy would be consists of these levels. So, here 1 A 1 G is very important 1 A 1 G is very important here at this state is greatly stabilized by the ligands and drops rapidly in energy as ligand field strength increases. So, A 1 G is here you can see 1 A 1 G and it drops here as we move from left to the right because they become low spin complex here it is a high spin complex here. It crosses the ground state 5 T 2 state and it becomes the ground state here 5 T 2 G was here and then it goes up and then A 1 G what happens that becomes ground state here. For example, if you look into COF 63 minus high spin blue in color 1 peak at 13000 centimeter minus 1 and then if you look into this one we have two transitions are there here we have two transitions here that is 1 A 1 G 1 A 1 G to 1 T 1 G the 1 T 1 G here and the other one is 1 A 1 G to 1 T 2 G here 1 T 2 G here. So, these two transitions are there whereas, here we will see only one transition case of COF 63 minus. So, you can very nicely identify from this D 6 thermal sugar non diagram for D 6 complexes such as hexafluorocobaltid 3 minus and tris ethylene diamine cobalt 3 plus low spin complex which shows two transitions. So, now let us look into ligand to metal charge transfer transitions. For example, I have shown here in case of a tetrahedral complex you can see here we have low energy field sigma orbital and low energy field pi orbital is there and then they cause this kind of low crystal free stabilization energy because both the electrons when they are donated to the metal what happens the homo-lomogap shrinks and then this is more destabilized. That is the reason most of the highlight complexes are very reactive and we use them very conveniently for doing substitution reactions to replace with better ligands. For example, if you take chlorocompones and if you add water it immediately forms hexaquacompone or if you add ammonia you can form hexamone compound or we can use any other ligand to replace very quickly because reactivity is more in these cases and they are all labile complexes. Then this is about metal to ligand charge transfer transition you can see here they have low energy field sigma orbital and high energy empty pi orbitals. Examples as I mentioned it can be CO or PR3 or aromatic groups, unsaturated aromatic groups double bond triple bond compounds and also pyridines etcetera. So, here because of the T2G electrons this essentially n non-bonding orbitals are T2G they combine with pi star to generate bonding and anti-bonding orbitals where this electrons will be coming and here due to this one the CFSC the homo-lomogap increases and they are more stabilized. So, ligand to metal charge transfer transition have very large extension coefficients. So, you can see here how this metal and ligand presence would have an impact on ligand to mass charge transfer transition. For example, first row and second row and third row how it varies it increases steadily and you go from 3D to 4D to 5D the gap increases and the energy required is very high or it falls into lower wavelength. So, now symmetry distortion one can see here trans diethylene diphyloro compound D3IR distorted from octahedral to D4H tetragonal elongation is there. When in tetragonal elongation what would happen is energy level of D3IR as the symmetry of its environment changes from octahedral to tetragonal and you can see these different transitions because of the change in the point group from OH to T4H here. Now, let us look into a couple of problems here very simple problems explain why an electronic transition for a high spin is spin formidable, but for cobalt 2 plus that is D7 is spin allowed. It is very simple you have to identify the oxygen state and the D system. So, in case of MN hexa aqua manganese 2 plus it is a D5 system and immediately of knowing you just write the electronic coefficient of splitting orbit crystal field splitting diagram and put the electrons you have all the five electrons are something like this high spin complex. So, it is spin formidable and it is very easy and then, but for cobalt it is a D7 system D7 system you have again high spin complex. So, here it is allowed because these two electrons can easily go to easy level. So, this is spin allowed transition it is very easy here. So, now let us look into what is the DN configuration spin multiplicity and term symbol of the ground state of titanium 3 plus and vanadium 3 plus already given here. So, this titanium 3 plus is a D1 system plus 3 state D1 system is there and then vanadium is D2 vanadium 3 plus means 3 D3 4 S2 it is a D2 system vanadium is D2 system and then, if you D1 system well you can write again you put one electron here. So, it is if you go for L value equals 0 1 2 3 S P D. So, it is a 2 D is here a spin multiplicity will be S equals half 2 S plus 1 will be 2 into half plus 1 it is 2 D. So, 2 D is shown here and similarly, if you go for vanadium D2 system 2 electrons here. So, L equals 3 that means, F and then S equals 1 therefore, 2 S plus 1 equals 3. So, this is 3. So, this is how the ground state term symbol I can spin multiplicity is 3 and 2 in this case and you can identify this. So, all these things can be done easily provided you identify the arson state in the metal complex and then find out what electronic configuration is there and simply find out L value S value and then J value if needed and then you should be able to write the term ground state term symbol. Now, the electronic spectrum of an aqueous solution of Triss ethylene diamine nickel 2 plus exhibits broad absorptions with lambda maximum at 325 550 and 900 nanometer. First question is suggest assignments for the electronic transitions the second one is which bands are in the visible region ok. So, for this one you can see here recollect the Argel diagram for D2 system or D8 system they are essentially same and then if you recall where exactly it comes you can see here 3 transitions will be there and these 3 transitions you can always write here and of course, here what happens the energy of this one drops here and then it increases here you consider like this. So, that does not matter here that question is not asked one we have to assign you should be able to assign once you identify the first ground term ground state and then 3 other states 1 due to P and 1 due to 2 due to F excluding the ground state of F system. So, you should be able to do that which bands are in the visible region. So, 900 nanometer assigned to 3A due to 3T2G and 550 is assigned to 3A2G to 3T1G and then 325 is assigned to 3A2G to 3T1G. So, this how you can identify and assign the values. So, of the 3 absorptions in this one which is closest to the UV end of the spectrum does the notation 3T3T2G to 3A2G indicate in absorption or emission band why are the 3 transitions spin all over and upward error these are the 3 questions you should work out. These questions I am posing to you people by just looking to the spectrum and then try to work out for these 3 questions. So, now one more problem is here aqueous solution of hexa-covanadium 3 plus shows absorptions at 17,000, 200 and 25,600 centimeters assigned to 2 transistors already given 3T1G F to 3T2G and then 3T1G F to 3T1G P transitions estimate values of B and delta O for hexa-covanadium compound and then here this is the compound given and then if you just look into here the plot E versus B versus delta versus B. So, here the B is unknown, but if you take the ratio what we get is E2 over E1, but if you consider E2 over E1 here this comes around E2 over E1 ratio will be 1.49. In order to get that 1.49 you have to do trial points here what you should do is you have to consider at different points for delta value with respect to this one considering the transition ground state as well as the excited state between which electronic transition happening, but that means 3T1G to 3T2G and 3T1G to 3T1G P. You have to consider this keep on doing by trial for example you when this value is 20 what would happen. So, then at this value what is the E2 by B or D1 by B for these two levels between each electronic transition taking place you can find out from the plot approximately and get the value here and similarly you make another trial considering 30 delta O over B equals 30 and then again at that value you find out E2 from y-axis and calculate these two values and get 1.46 and then one more trial you do it and you arrive at the one we found from the data. So, that means there is an approximate answer, but we are now be able to estimate B and delta as follows you can take here 29 we have in this one what we are getting is 40 and then E2 equals 25600 and B equals 640 and then when you take 29 here again E1 by B you can take it at 26.9 is there since E1 equals 17200 B equals 640 centimeter inverse. So, substitution of the value of B into this one would give an estimate of delta this is approximately 18600 centimeter minus 1 this how you should be able to do from this taking simply Ternobetsugano diagram. So, this shows how useful Ternobetsugano diagram from which by simply looking into the transition and comparing the value of E by B versus delta O by B and you can calculate delta O very easily for a given complex. So, let me stop here and come with more examples maybe at the end in my next lecture I shall focus your attention on IR spectroscopy after that one I will go to mass spectrometry and then EPR and if time permits mass bar and then I would come back to solve problems from all these spectroscopic methods to make you very expert in elucidation and interpretation of the data. So, until then see you until my next lecture.